A telephone wire stretched tight between two poles placed ten meters apart is a favorite resting spot for a flock of crows.

Suppose one morning two crows land on the wire, each at a random spot (the probability is uniformly distributed). With a bucket of paint and a brush you mark the stretch of wire between them. A certain length of wire will have been painted.

On average, what length of wire would you expect to have painted? Assume that each bird is a single point along the line, and so has no width.

Suppose instead that a dozen crows landed on the wire, each at an independent, random location, and you painted the stretch of wire between each bird and its nearest neighbor. On average, what total length of wire would you expect to have painted now?

And if a thousand crows landed?

A computer-generated solution could be found, but bonus points will be awarded for a formal proof!

(In reply to re: Some thoughts on the 2 bird problem. (partial solution) by Bob)

If we paint from the first pole to a sigular bird on the wire, then the average is 1/2.

As stated, the average space between 2 birds approaches 1/3.

With 3 birds, would the average be higher than 1/2?

Does this problem have an oscillation between even and odd numbers of birds?  It seems that for even numbered birds there would be cases where the birds clump in pairs, but with odd numbered birds there would always be 'the odd bird out', who supplies one more space to be painted which wouldn't otherwise get painted when there is an even number.

Posted by Erik O. on 2004-06-07 14:42:22